Expert: Socrates - 10/22/2007

Question
A conical water tank with vertex down has a radius of 10 ft at the top and is 24 ft. high. if water flows into the tank at a rate of 20ft^3/ min., how fast is the depth of the ater increasing when the water is 16 ft. deep?

Answer
Let r be the radius of the circle of water at the top , h the height or depth of the water and V the volume of water in the cone. Of course, r, h and V are functions of time.

We have V = (1/3)(pi)r^2 h , the formula for the volume of a cone.

Then V' = (2/3)(pi)rr'h + (1/3)(pi)r^2 h'

V' is the rate that the volume is increasing, so V' = 20
We are looking for h' , the rate that the height or depth is increasing when h = 16 . Because the tank is a cone , r/h = 10/24 = 5/12 , so when h=16 , r/16 = 5/12 and
r = 20/3 . Since r/h = 5/12 , r = (5/12)h and r' = (5/12)h'.

So far , we have

V' = 20
h = 16
r = 20/3
r' = (5/12)h'

at the time when the water is 16 ft deep

Substitute these values into


V' = (2/3)(pi)rr'h + (1/3)(pi)r^2 h'

getting

20 = (2/3)(pi)(20/3)(5/12)h'(16) + (1/3)(pi)(400/9) h'

20 = (3200/108)(pi)h' + (400/27)(pi)h'

20 = (800/27)(pi)h' + (400/27)(pi)h'

20 = (1200/27)(pi)h'

20 = (400/9)(pi)h'

(9/20) = (pi)h'

h' = 9/(20 pi )

when the water is 16 ft deep , the depth is increasing at
9/(20 pi) feet per minute